I need to find all of the zeros.
g(x) = x^4 + x^3 -2x^2 +6x -4
Oh sh** sorry I thought you had written "quadratics", but you indeed wrote "quartics". I'm out, too hard for me ...
But maybe Wikipedia could help ?
http://en.wikipedia.org/wiki/Quartic_function
So the bad new is that is has no linear factors. We can check for linear factors by using the rational roots theorem.
So It must factor into two quadratics
since the lead term has coeffient 1 we can write the two quadratics as
$\displaystyle (x^2+ax+b)\cdot (x^2+mx+n) =x^4+x^3-2x^2+6x-4$
Now we multiply out the left hand side to get
$\displaystyle x^4+(m+a)x^3+(n+am+b)x^2+(an+bm)x+bn=x^4+x^3-2x^2+6x-4$
This give the system of equations that is hard to solve and requires alot of trial and error.
$\displaystyle m+a=1,n+am+b=-2,an+bm=6,bn=-4$
Now comes the hard work solving this system of equations.
If $\displaystyle b=2,n=-2$ then if we plug into the 2nd equation
$\displaystyle am=-2$ this gives two choices for a and m
The correct choice is $\displaystyle a=-1,m=2$ they both check in equation 3.
Now plugging back into the equation from the beginning we get
$\displaystyle (x^2-x+2)\cdot (x^2+2x-2) $
It's impossible to factor quartics... just kidding, but I can't offer you a better method than guess and check, I mean I could, but I don't think it's any better, so....
Quartics can be factored into two quadractics (provided they have real coefficients)
$\displaystyle g(x) = x^4 + x^3 -2x^2 +6x -4$
$\displaystyle g(x)=(x^2+\quad )(x^2+\quad )$
Now I start playing around with stuff, we need a $\displaystyle +x^3$ and a $\displaystyle -2x^2$
Well it looks like $\displaystyle -x$ and $\displaystyle +2x$ will do the job
Since $\displaystyle x^2(-x)+x^2(2x)=x^3$ and $\displaystyle (-x)(2x^2)=-2x^2$
So $\displaystyle g(x)=(x^2-x+\quad )(x^2+2x+\quad )$
Now the final number is -4, so our choices consist of 1,2,4 with necessary signs out in front, just running through those, again I swear this is all guess and check and it's not necessarily quick, you'll get 2 and -2
$\displaystyle g(x)=(x^2-x+2)(x^2+2x-2)$
There are formulas and things you can look up and use, but this question comes up so rarely for me that I have to stumble through it. It's not an efficient way of doing it.